Question

A study considered a sample of 50 observations used to predict SALES. Included in the analysis were 9 predictors variables, ( Independent Variables).

	X 1	X 2	X 3	X 4	X 5	X 6	X 7	X 8	X 9
X 2	0.804
X 3	0.625	0.443
X 4	0.032	0.032	0.231
X 5	0.159	0.214	0.177	-0.194
X 6	0.319	0.373	0.308	0.054	0.293
X 7	-0.016	0.030	0.079	0.168	-0.309	0.067
X 8	-0.026	0.103	0.015	0.151	-0.311	0.059	0.912
X 9	0.169	-0.027	-0.104	0.017	-0.248	0.114	0.174	0.223
SALES	0.764	0.630	0.756	0.149	0.171	0.426	0.145	0.141	-0.068

1. Based on the correlation matrix shown below, is there any concern about Multicollinearity?
   - If YES, list all pairs of variables involved:
   
   -What limit did you use?
   
   - What action must be taken to eliminate multicollinearity?  Be very specific.  
   
   -Which variable appears to have the weakest relationship with SALES?
   
   -The following is the initial Regression Printout. Variable X2, and X8 were not included in the regression. Why?  
   

   Source	DF	Adj SS	Adj MS	F-Value
   Regression	7	145157	20736.7	19.62
   Error	42	44385	1056.8
   Total	49	189543

   
   
   Use the F test and a 0.05 level of significance to determine whether the Regression model is significant.
   
   What is the value of the F Critical Point
   
   How many degrees of freedom did you use?
   
   Compute the R-Square:
   
   Show your work and compute the Adjusted R-Square:

Answer 1

Based on the correlation matrix shown below, is there any concern about Multicollinearity?

Yes , there any concern about Multicollinearity

For Pair ( X7 , X8 ) the correlation is 0.912 , which implies they are highly correlated with each other.

Also Pair ( X1 ,X2 ) has correlation = 0.804 , so these pair is also highly correlated

So this pairs of variables involved : Pair ( X1, X2) and Pair ( X7, X8 )

-What limit did you use?
Limit we have used is " > | 0.70 | " for any pair

- What action must be taken to eliminate multicollinearity?  Be very specific.  

The only think can be done is to remove any of one variables of each pair which have lowest relationship with SALES .

Like in pair ( X7 , X8 ) X8 has lowest relationship with SALES ( which is 0.141 ) , so it is better to remove it .

-Which variable appears to have the weakest relationship with SALES?
X9 variable have the weakest relationship with SALES ( which is -0.068 )

-The following is the initial Regression Printout. Variable X2, and X8 were not included in the regression. Why?  

Source	DF	Adj SS	Adj MS	F-Value
Regression	7	145157	20736.7	19.62
Error	42	44385	1056.8
Total	49	189543

Use the F test and a 0.05 level of significance to determine whether the Regression model is significant.

To test

H0 : bi = 0 , i = 1,3,4,5,6,7,9    { Model is not significant }

H1 : bi 0    for atleast one i    { Model is significant }

Test Statistics F:

F = MS_R / MS_RES = 20736.7 / 1056.8 = 19.62216

Thus calculated F- value is 19.62

What is the value of the F Critical Point

It is given by

is F-distributed with df1= 7 and df2=42 degree of freedom and =0.05,

It can be computed from statistical book or more accurately from any software like R,Excel

From R

> qf(1-0.05,df1=7,df2=42)
[1] 2.23707

Thus value of the F Critical Point is 2.23707

We reject null hypothesis if calculated F-value is less than

How many degrees of freedom did you use?

is F-distributed with df1= 7 and df2=42 degree of freedom and =0.05,

Conclusion -

Since F- value = 19.62 > 2.23707 i.e F- value >

So we reject null hypothesis at 5% of level of significance at hence conclude that model is significant.

Compute the R-Square:

Formula :

R-Square = 1 - SS_RES / TSS

                = 1 - 44385 / 189543

            = 0.7658315

Thus R-Square = 0.7658315

Show your work and compute the Adjusted R-Square:

Formula :

Adjusted R-Square: = 1 - SS_RES/ df(Error) / TSS / df(Total)

        or = 1 - SS_RES/ (n-k ) / TSS / (n-1)         

Here k = 7        ( number of regressor )

also   n-k = 42    ( given)     

               = 1 - ( 44385 / 42 ) / ( 189543 / 49 )

               = 0.7268034

Thus Adjusted R-Square = 0.7268034